Question

please do it urgently

Answer 1

1. wt of litharge = 10.78 g
      Wt of lead    = 10 g
     Wt of oxygen = 0.78 g

     in litharge wt of oxygen that combines with 1 g of lead = 0.078 g

2. Wt of red lead = 9.775 g
   Wt of litharge    =  9.545 g
10.78 g of litharge contains 10 g of lead
therefore 9.545 g of litharge contains  10 x 9.545 / 10.78 = 8.854 g of lead

wt of oxygen in 9.775 g of red lead = 9.775 - 8.854 = 0.921 g oxygen

therefore 1 g of lead in combination with 0.921/8.854 = 0.104 g of oxygen

3. Wt of lead peroxide = 4.87 g
    wt of litharge =  4.545 g

10.78g  of litharge contains 10 g lead
therefore 4.545 g of litharge contains 10 x 4.545 / 10.78 = 4.216 g of lead

wt of oxygen in 4.87 g of lead peroxide = 4.87 - 4.216 = 0.654 g oxygen
therefore 1 g of lead in lead peroxide combines with 0.654/4.216 = 0.155 g of oxygen.

The ratio of oxygen that combines with 1 g of lead in litharge, red lead and lead peroxide is

0.078:0.104:0.155
1: 4/3: 2
 or

3:4:6

This illustrates the law of  multiple proportions